Find Derivative: Solve G'(3) Problem

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The discussion focuses on calculating the derivative G'(3) for the function G(x) defined as G(x) = 6f(x) - g(x). The derivative is expressed as G'(x) = 6f'(x) - g'(x), leading to G'(3) = 6f'(3) - g'(3). Participants determined that f'(3) = -3 and g'(3) = 1, resulting in G'(3) = 6(-3) - 1 = -19.

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View attachment 6132

Does it want me to plug in G'(3) into the equation?
 

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We are told:

$$G(x)=6f(x)-g(x)$$

And so we know:

$$G'(x)=6f'(x)-g'(x)$$

which means:

$$G'(3)=6f'(3)-g'(3)$$

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D
 
MarkFL said:
We are told:

$$G(x)=6f(x)-g(x)$$

And so we know:

$$G'(x)=6f'(x)-g'(x)$$

which means:

$$G'(3)=6f'(3)-g'(3)$$

So, what you need to do is determine from the graph the values of $f'(3)$ and $g'(3)$, and plug them into the above...:D
will it be -26 when G'(3)
 
the said:
will it be -26 when G'(3)

What values do you get from the graph for $f'(3)$ and $g'(3)$?
 
MarkFL said:
What values do you get from the graph for $f'(3)$ and $g'(3)$?

you get f'(3) = 6 and g'(3) = 1
 
Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.
 
MarkFL said:
Around $x=3$, observe that for every increase of 1 unit in $x$, we find $f$ decreases by 3 units...thus $f'(3)=-3$, and I agree that $g'(3)=1$.

how did you get -3 from f'(3)?
 
the said:
how did you get -3 from f'(3)?

Rise over run...rise is -3 when run is 1...:D
 

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